Question: You have found the following ages (in years) of all 4 zebras at your local zoo: $ 8,\enspace 13,\enspace 5,\enspace 10$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{8 + 13 + 5 + 10}{{4}} = {9\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $-1$ years $1$ year $^2$ $13$ years $4$ years $16$ years $^2$ $5$ years $-4$ years $16$ years $^2$ $10$ years $1$ year $1$ year $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1} + {16} + {16} + {1}} {{4}} $ $ {\sigma^2} = \dfrac{{34}}{{4}} = {8.5\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{8.5\text{ years}^2}} = {2.9\text{ years}} $ The average zebra at the zoo is 9 years old. There is a standard deviation of 2.9 years.